# Starting with #\tt{SO_3(g)}# at #\sf{"1.00 atm"}#, what will be the total pressure when equilibrium is reached with the following reaction at #\sf{"700 K"}#?

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#\tt{2SO3(g)\harr2SO_2(g)+O_2(g)}#

#\sf{K_p=1.6xx10^-5" @ 700 K"}#

(Has this been posted before? I apologize if yes)

(Has this been posted before? I apologize if yes)

##### 1 Answer

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Well, let's think about what we want at the end. As in **Dalton's law of partial pressures** for ideal gases,

#P_("tot") = P_(SO_3) + P_(SO_2) + P_(O_2)#

All of these quantities are equilibrium quantities. Once we know all three (actually, we will find that we only need one quantity), we know how to find the total pressure.

** Now** it is practical to write the

**ICE table**, since we have an idea of what path to follow. In terms of

#2"SO"_3(g) " "rightleftharpoons" " 2"SO"_2(g) + "O"_2(g)#

#"I"" "1.00" "" "" "" "" "0" "" "" "" "0#

#"C"" "-2x" "" "" "" "+2x" "" "+x#

#"E"" "1.00-2x" "" "" "2x" "" "" "x#

Now what we find is that the **total pressure** is given by:

#P_"tot" = (1.00-2x) + (2x) + (x) = 1.00 + x#

And so, if we find

#1.6 xx 10^(-5) = (P_(SO_2)^2P_(O_2))/(P_(SO_3)^2)#

#= ((2x)^2x)/(1.00-2x)^2#

We can tell that

#1.6 xx 10^(-5) ~~ (4x^3)/(1.00)^2#

Thus,

#x ~~ ((1.6 xx 10^(-5) cdot (1.00)^2)/4)^(1//3)#

#=# #0.015_9# #"atm"# to two sig figs.

and so,

#color(blue)(P_"tot") = "1.00 atm" + 0.015_9 # #"atm"#

#=# #color(blue)("1.02 atm")# to two decimal places.